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22-Mec-B8 Engineering Materials · May 2017

Question 7 of 8: E-glass/PVC composite — modulus, load sharing and strain

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, May 2017 — 16-Mec-B8 Engineering Materials. Three hours, open book; any non-communicating calculator permitted. Eight problems, all of equal value; any FIVE constitute a complete paper, so each problem is worth 20 marks. Candidates are urged to submit a clear statement of any assumptions made. All eight problems are solved below, because the set as a whole is the study resource.

Reference texts (22-Mec-B8 Engineering Materials).

  • Askeland & Wright, The Science and Engineering of Materials, 7th ed. — the primary syllabus text.
  • Callister & Rethwisch, Materials Science and Engineering: An Introduction, 10th ed.
  • Dieter, Mechanical Metallurgy, 3rd ed. — Considère's construction and plastic instability.
  • Shackelford, Introduction to Materials Science for Engineers, 8th ed. — ceramics, glasses and glass-ceramics.
  • Fontana, Corrosion Engineering, 3rd ed. — galvanic series, sacrificial protection and Faraday's law.
  • Ashby, Materials Selection in Mechanical Design, 5th ed. — selection criteria and material indices.
  • Polmear, Light Alloys, 5th ed. — aluminium tempers, Al–Li alloys and maraging steels.

Note on this sitting. Seven of the eight problems restate archetypes seen in 2013–2016 with fresh data, and every calculation below has been re-worked from this paper's numbers. Question 8 is the genuinely new one: the familiar aluminium–lithium substitution has been expanded into a three-alloy selection problem that also asks for the beam volume and for a strength-to-density ranking, and the two halves of it do not pick the same alloy.

Question 7: E-glass/PVC composite — modulus, load sharing and strain (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A continuous, aligned E-glass/PVC composite loaded along the fibre direction:

Given data
QuantitySymbolValue
Modulus of E-glass fibreEf75 GPa
Modulus of hardened PVC matrixEm2.3 GPa
Volume fraction of PVC matrixVm0.72
Volume fraction of glass fibreVf0.28
Cross-sectional areaA600 mm2
Longitudinal loadP73 000 N

Find. (a) the longitudinal modulus of the composite, (b) the percentage of the applied load carried by the fibres, and (c) the axial strain under the stated load.

unidirectional laminate loaded along the fibres — equal strain in both phasesE-glass Vₜ = 0.28 | PVC Vₘ = 0.72load share92.7 %7.3 % matrixE_c = 22.656 GPa (rule of mixtures)
Continuous aligned composite loaded along the fibres. Both phases are bonded and stretch together, so the strain is common; the stiffer phase therefore carries the higher stress, and the load divides in proportion to EiVi.

Approach. Loading along continuous aligned fibres is the isostrain (Voigt) case: fibre and matrix suffer the same strain, so the composite modulus is the volume-weighted average of the two moduli and the load divides in proportion to the product of modulus and volume fraction. The strain then follows from the composite stress and the composite modulus.

  1. Fix the volume fractions. The matrix occupies 72 % of the volume, so the reinforcement occupies the rest: $$V_f = 1 - V_m = 1 - 0.72 = 0.28$$ It is worth pausing on this line, because the question quotes the matrix fraction while every formula that follows is written in terms of the fibre fraction.
  2. Part (a) — apply the rule of mixtures for the longitudinal modulus. In the isostrain condition the composite modulus is $$E_c = E_fV_f + E_mV_m = 75(0.28) + 2.3(0.72) = 21.00 + 1.656$$ $$\boxed{\ E_c = 22.656\ \text{GPa}\ }$$ Almost 93 % of that stiffness comes from the 28 % of the volume that is glass, which is the entire point of reinforcing at all.
  3. Part (b) — partition the load between the phases. Since both phases carry the same strain ε, the stress in each is σi = Eiε, and the force in each is that stress times its share of the area, which for aligned fibres equals its volume fraction. Hence $$\frac{P_f}{P_c} = \frac{E_fV_f\varepsilon}{(E_fV_f + E_mV_m)\varepsilon} = \frac{E_fV_f}{E_c} = \frac{21.00}{22.656}$$ $$\boxed{\ \frac{P_f}{P_c} = 0.9269\ \text{, i.e. } 92.69\ \% \text{ carried by the fibres}\ }$$ leaving only 7.31 % to the PVC. The strain cancels out of that ratio, which is why the load split is a property of the material and not of the load applied to it.
  4. Part (c) — find the composite stress under the stated load. Referring the load to the whole section, $$\sigma_c = \frac{P}{A} = \frac{73\,000\ \text{N}}{600\ \text{mm}^2} = 121.67\ \text{MPa}$$
  5. Obtain the strain from Hooke’s law for the composite. With Ec = 22.656 GPa = 22 656 MPa, $$\varepsilon = \frac{\sigma_c}{E_c} = \frac{121.67}{22656}$$ $$\boxed{\ \varepsilon = 5.3702\times 10^{-3} = 0.5370\ \%\ }$$
  6. Check the answer against the phase stresses. A useful closing check: at this common strain the fibre stress is σf = 75 000(0.005370) = 402.8 MPa and the matrix stress is σm = 2300(0.005370) = 12.35 MPa. Recombining them over their areas returns 402.8(0.28)(600) + 12.35(0.72)(600) = 67 664 + 5 336 = 73 000 N, the applied load, and the fibre share 67 664/73 000 = 92.7 % confirms part (b).
Results for the E-glass/PVC composite
QuantitySymbolValue
Fibre volume fractionVf0.28
(a) Longitudinal modulus of the compositeEc22.656 GPa
(b) Share of the load carried by the glassPf/Pc92.69 %
Share carried by the PVC matrixPm/Pc7.31 %
Composite stress under 73 kNσc121.67 MPa
(c) Axial strain under 73 kNε5.3702 × 10−3 (0.5370 %)

Check: the solution assumes continuous, perfectly aligned fibres with a perfect fibre–matrix bond, loading along the fibre axis, and both phases still elastic. Discontinuous or misaligned fibres require a length-efficiency factor, and a transverse load would call for the isostress (Reuss) rule of mixtures, which here would give a composite modulus of only 3.157 GPa — about one seventh of the longitudinal value. This anisotropy is the defining feature of an aligned composite, not a defect in the calculation. Note also that at 0.54 % strain the PVC is at the edge of, or a little beyond, its own linear range, so a real design check would use the matrix secant modulus rather than the initial 2.3 GPa.